All categories

3 years ago

Help me with 3, and 5 show your work !

Mathematics

1 answer:

ss7ja [257]3 years ago

5 0

Okay so for number 3, you have to do the top work first.

3: POSITIVE 2!! first, you do the stuff inside the parentheses first because of P(parentheses)EMDAS. so, 14-2-10! -2-10 is -12 and then plus positive 14 is +2. but, the negative sign outside of the parentheses makes that +2 a -2.
but, you cant forget the -12 outside. you have to do -12-2 which gets you -14. then, this is easy! -14 divided by -7 is a positive 2!

5: POSITIVE 3!! again, do the stuff in the parentheses first!! -2-4 is -6. then, -6 x 2 is -12! so, divide -12 by -4 and you get a positive 3!

You might be interested in

Which term is used to describe a number that makes a statement true?

frosja888 [35]

That is called the solution

factor is like 2 and 3 and 9 are factors of 18 because 2*9=18 and 3*6=18

inequality is like 3<4

product is multiplcation so like product of x and y is x times y

answer is solution

8 0

3 years ago

Location is known to affect the number, of a particular item, sold by an automobile dealer. Two different locations, A and B, ar

yKpoI14uk [10]

Answer:

We conclude that the true mean number of sales at location A is fewer than the true mean number of sales at location B.

Step-by-step explanation:

We are given that Location A was observed for 18 days and location B was observed for 13 days.

On average, location A sold 39 of these items with a sample standard deviation of 8 and location B sold 49 of these items with a sample standard deviation of 4.

Let $\mu_1$ = true mean number of sales at location A.

$\mu_2$ = true mean number of sales at location B

So, Null Hypothesis, $H_0$ : $\mu_1-\mu_2\geq0$ or $\mu_1 \geq \mu_2$ {means that the true mean number of sales at location A is greater than or equal to the true mean number of sales at location B}

Alternate Hypothesis, $H_A$ : $\mu_1-\mu_2$ or $\mu_1< \mu_2$ {means that the true mean number of sales at location A is fewer than the true mean number of sales at location B}

The test statistics that would be used here Two-sample t test statistics as we don't know about the population standard deviations;

T.S. = $\frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }$ ~ $t_n__1_-_n__2-2$

where, $\bar X_1$ = sample average of items sold at location A = 39

$\bar X_2$ = sample average of items sold at location B = 49

$s_1$ = sample standard deviation of items sold at location A = 8

$s_2$ = sample standard deviation of items sold at location B = 4

$n_1$ = sample of days location A was observed = 18

$n_2$ = sample of days location B was observed = 13

Also, $s_p=\sqrt{\frac{(n_1-1)s_1^{2}+(n_2-1)s_2^{2} }{n_1+n_2-2} }$ = $\sqrt{\frac{(18-1)\times 8^{2}+(13-1)\times 4^{2} }{18+13-2} }$ = 6.64

So, test statistics = $\frac{(39-49)-(0)}{6.64 \times \sqrt{\frac{1}{18}+\frac{1}{13} } }$ ~ $t_2_9$

= -4.14

The value of t test statistics is -4.14.

Now, at 0.01 significance level the t table gives critical value of -2.462 at 29 degree of freedom for left-tailed test.

Since our test statistics is less than the critical values of t as -2.462 > -4.14, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which we reject our null hypothesis.

Therefore, we conclude that the true mean number of sales at location A is fewer than the true mean number of sales at location B.

3 0

3 years ago

A very large data set (N > 10,000) has a mean value of 1.65 units and a standard deviation of 72.26 units. Determine the rang

Romashka [77]

Answer:

$z=-0.674$

And if we solve for a we got

$a=1.65 -0.674*72.26=-47.05$

$z=0.674$

And if we solve for a we got

$a=1.65 +0.674*72.26=50.35$

So then the limits where 50% of the data lies are -47.05 and 50.35

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

Solution to the problem

Let X the random variable that variable of interest of a population, and for this case we know the distribution for X is given by:

$X \sim N(1.65,72.26)$

Where $\mu=1.65$ and $\sigma=72.26$

For this case we want the limits for the 50% of the values.

So on the tails of the distribution we need the other 50% of the data, and on ach tail we need to have 25% since the distribution is symmetric.

Lower tail

For this part we want to find a value a, such that we satisfy this condition:

$P(X>a)=0.75$ (a)

$P(X$ (b)

Both conditions are equivalent on this case. We can use the z score again in order to find the value a.

As we can see on the figure attached the z value that satisfy the condition with 0.25 of the area on the left and 0.75 of the area on the right it's z=-0.674. On this case P(Z<-0.674)=0.25 and P(z>-0.674)=0.75

If we use condition (b) from previous we have this:

$P(X$

$P(z$

But we know which value of z satisfy the previous equation so then we can do this:

$z=-0.674$

And if we solve for a we got

$a=1.65 -0.674*72.26=-47.05$

So the value of height that separates the bottom 25% of data from the top 75% is -47.05.

Upper tail

For this part we want to find a value a, such that we satisfy this condition:

$P(X>a)=0.25$ (a)

$P(X$ (b)

Both conditions are equivalent on this case. We can use the z score again in order to find the value a.

As we can see on the figure attached the z value that satisfy the condition with 0.75 of the area on the left and 0.25 of the area on the right it's z=0.674. On this case P(Z<0.674)=0.75 and P(z>0.674)=0.25

If we use condition (b) from previous we have this:

$P(X$

$P(z$

But we know which value of z satisfy the previous equation so then we can do this:

$z=0.674$

And if we solve for a we got

$a=1.65 +0.674*72.26=50.35$

So the value of height that separates the bottom 75% of data from the top 25% is 50.35.

So then the limits where 50% of the data lies are -47.05 and 50.35

6 0

3 years ago

Evaluate. (-5)5 Enter your answer in the box.

Klio2033 [76]

Answer:

The answer is -25

Step-by-step explanation:

7 0

3 years ago

What term describes a poll often used to find the opinions of a group of people?

KengaRu [80]

I would go with

Surveys

Because it's a great way to get the opinions of a lot of people

Hope that helps :)

8 0

3 years ago